Question

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. Block 1 has a mass of 2.25 kg and is on an incline of θ₁=46.5° with coefficient of kinetic friction μ₁=0.205 . Block 2 has a mass of 7.25 kg and is on an incline of θ₂=35.5° with coefficient of kinetic friction μ₂=0.105 . The two-block system is in motion with the block of mass 2 sliding down the ramp. Find the magnitude of the acceleration of 2 down the incline.

Answer 1

1. Block 1 (2.25 kg) on the incline with friction:
- Weight of block 1 (mg₁): 2.25 kg * 9.81 m/s² = 22.0725 N
- Normal force on block 1 (N₁): 2.25 kg * 9.81 m/s² * cos(46.5°) = 14.851 N
- Frictional force on block 1 (f₁): μ₁ * N₁ = 0.205 * 14.851 N = 3.044255 N
- Net force on block 1 (F₁): mg₁ * sin(46.5°) - f₁ = 22.0725 N * sin(46.5°) - 3.044255 N = 13.223 N

2. Block 2 (7.25 kg) on the incline with friction:
- Weight of block 2 (mg₂): 7.25 kg * 9.81 m/s² = 71.2725 N
- Normal force on block 2 (N₂): 7.25 kg * 9.81 m/s² * cos(35.5°) = 58.104 N
- Frictional force on block 2 (f₂): μ₂ * N₂ = 0.105 * 58.104 N = 6.10032 N
- Net force on block 2 (F₂): mg₂ * sin(35.5°) - f₂ = 71.2725 N * sin(35.5°) - 6.10032 N = 41.393 N

Now, we can set up equations of motion for both blocks using Newton's second law (F = ma), where a is the acceleration:

For block 1:
F₁ = m₁ * a
13.223 N = 2.25 kg * a₁

For block 2:
F₂ = m₂ * a
41.393 N = 7.25 kg * a₂

Now, we can solve for a₂ (the acceleration of block 2):
a₂ = 41.393 N / 7.25 kg = 5.704 m/s²

So, the magnitude of the acceleration of block 2 down the incline is approximately 5.704 m/s².